Take two graphs (of bounded valency) or manifolds (f bounded geometry) $G$ and $G'$. Assume there is a quasi-isometry $f:G \to G'$, and assume the Poisson or Martin boundary of $G$ is known, what may one say about the Poisson or Martin boundary of $G'$ (or how does $f$ transforms Poisson or Martin boundaries)?

EDIT: a map $f:X \to Y$ between two metric spaces $(X,d_X)$ and $(Y,d_Y)$ is a quasi-isometry if $\exists C \geq 0$ and $D \geq 1$ such that

Not an answer, unfortunately. It is known that the Martin boundary dominates the geodesic boundary (Prop.I.7.15 of Borel & Lizhen, Compactifications of symmetric and locally symmetric spaces). So if quasi-isometry does not preserve the geodesic boundary then it does not preserve the Martin one either. Points of the geodesic boundary are equivalence classes of (unit speed, I think) geodesics $\gamma(t)$ such that $\limsup_{t\to\infty} d(\gamma_1(t),\gamma_2(t)) < +\infty$. How does a quasi-isometry act on geodesics?
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Igor KhavkineSep 13 '12 at 9:13

Thanks for the reference in any case! A quasi-isometry would send those two geodesics at bounded distance but they might no longer be geodesics. Perhaps there are examples where the geodesic boundary is not preserve... I'll have to look into this.
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AntoineSep 13 '12 at 10:06

1

Even existence of nonconstant positive harmonic functions is not invariant under quasi-isometries.
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MishaSep 13 '12 at 11:08

@Misha, I think that is less of an obstacle if the Martin boundary is constructed using positive superharmonic functions. But that is an issue for the precise set of hypothesis that the OP desires.
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Igor KhavkineSep 13 '12 at 11:31

2

The reference is Terry Lyons paper in Journal of Diff Geometry, 1987. Igor: Geodesics do not map close to geodesics unless you are in Gromov hyperbolic setting. This fails already for Euclidean plane. Moreover, there examples of Croke and Kleiner where topological type of CAT(0) boundary is not preserved by quasi-isometries.
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MishaSep 13 '12 at 12:08

It is well-known that, in general, even for CAT(0) spaces, quasi-isometries do not induce maps of geometric boundaries. The easiest example to consider is the case of quasi-isometries of the Euclidean plane ${\mathbb R}^2$. Let $f(x)$ be any $L$-Lipschitz function ($L\ge 1$) of one variable (you can take, for instance, $f(x)=|x|$). Define a self-map $F(x,y)= (x, f(x)+y)$ of ${\mathbb R}^2$. Then $F$ is an $2L$-bilipschitz homeomorphism of ${\mathbb R}^2$. Considering images of the $x$-axis under $F$, we see that they are (in general) not close to geodesics (say, the graph of $y=|x|$). Moreover, Croke and Kleiner constructed example ("Spaces with nonpositive curvature and their ideal boundaries", Topology, 2000) of a CAT(0) group $G$ so that $G$ acts geometrically (discretely, isometrically, cocompactly) on two CAT(0) spaces $X, X'$ so that the geometric boundaries of $X, X'$ are not homeomorphic. Existence of geometric action of $G$ on $X, X'$ implies that these spaces are quasi-isometric to each other.

Still, for some interesting classes of spaces, quasi-isometries act naturally on all three boundaries (geometric, Poisson, Martin); this happens for instance, in the case of Cayley graphs of hyperbolic groups.

I believe Lyons references an earlier paper where the first example of quasi isometric Riemannian manifolds with non-homeomorphic Martin boundaries was established. Also, I think your point 3 is interesting, and it would be cool to know of a list of conditions that guarantee a good behavior of boundaries under quasi isometries.
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Gjergji ZaimiSep 14 '12 at 1:21

Gjergji: Yes, he does mention it, I forgot the reference though, I think it was unpublished at the time. I will add few examples where geometric boundary is preserved by quasi-isometries.
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MishaSep 14 '12 at 1:34

It is well know that the Poisson boundary of simple random walk on graph is not invariant under quasi isometries. Here's a construction:

Take $\Z^4$ and notice that (1) Its Poisson boundary is trivial (i.e. it is Liouville) and (2) a random walk starting anywhere on the line $L=\{(n,0,0,0)\mid n\in\Z \}$ has positive probability to never hit this line.

Let $\T=\{0,1\}^*$ be the infinite binary tree and let $S$ be the set of balanced vertices, that is, vertices with the same number of 0s and 1s in their name. Connect the vertices of $S$ with those of $L$ in an arbitrary 1-1 manner. Call the resulting graph $G$.

Now, a random walk on $G$, started anywhere in the tree, will hit $S$ almost surely. When it does, there is a positive probability for it to be absorbed in the $\Z^4$ part (i.e. to stay there forever). If this doesn't happen, then the walk will just hit $S$ again and again until it is absorbed. Since $\Z^4$ was Liouville, we get that $G$ is also Liouville.

However, consider the graph $G'$ which is identical to $G$ except that we replace each right going edge of the tree $\T$ with a path of length 2. This new graph is clearly quasi-isometric to $G$, but now there is a positive probability that the random walk will never hit $S$ and stay in the tree part of the graph forever. This give rises to a non-constant, bounded harmonic function $f(v)$ which is simply the probability of a random walk started at $v$ to be absorbed in $\Z^4$.